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You have been assigned to design a heat exchanger to cool a gaseous process stream in a chemical plant. The stream (20 kg/s) enters at 431 K, has a heat capacity of 3.45 J/g degree C, and needs to be cooled to 402 K. Cooling water is available at 85degreeF and has a specified maximum temperature of 120degreeF. The overall heat-transfer coefficient is approximately 570 W/m2 K. How much area must the heat exchanger have?

Explanation / Answer

Temperature is same in both units Kelvin and Degree celsius.

Given:

mass flow rate of gaseous stream, m = 20 kg/s

heat capacity of gas, Cp = 3.45 J/g deg C = 3.45*1000 J/kg deg C

Overall heat transfer coefficient, h = 570 W/(m^2 * K ) or h = 570 W/(m^2 * deg C )

Inlet temperature of hot stream, TH1 = 431 K = 157.8 C

Outlet temperature of hot stream, TH2 = 402 K = 128.8 C

Inlet temperature of cold water, TCW1 = 85 F = 29.4 C

Outlet temperature of cold water, TCW2 = 120 F = 48.9 C

Step1: Logarithmic mean temperature, LMTD = (deltaT1 - deltaT2)/(ln(deltaT1/deltaT2))

deltaT1 = TH1 - TCW2 = ( 157.8 C - 48.9 C ) = 108.9 C

deltaT2 = TH2 - TCW1 = ( 128.8 C - 29.4 C ) = 99.4 C

LMTD = ( 108.9 C - 99.4 C ) / ln ( 108.9 / 99.4 )

LMTD = 104.1 deg C

Step 2: Rate of heat transfer required, Q = m*Cp*(change in temperature of stream)

change in temperature of stream = 431 K - 402 K = 29 K = 29 deg C

So, Rate of heat transfer required, Q = m*Cp*(change in temperature of stream)

= (20 kg/s)*(3450 J/kg deg C)*( 29 deg C)

= 20*3450*29 ( kg/s)*(J/kg deg C)*(deg C)

= 2001000 J/s

= 2001000 W = 2.001* 10^6 W

Step 3:Design equation, Q = h*A*LMTD

2.001* 10^6 W = (570 W/(m^2 * deg C) )*A*(104.1 deg C)

A = 2.001* 10^6 W/ { (570 W/(m^2 * deg C) )*(104.1 deg C) }

= 2.001* 10^6/(570*104.1 ) m^2

A = 33.722m^2

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